# Find confidence levels from given intervals. A set of 40 data items, produces a confidence interval for the population mean of 94.93<mu<105.07. If sum x^2=424375, find the confidence level.

Find confidence levels from given intervals
A set of 40 data items, produces a confidence interval for the population mean of $94.93<\mu <105.07$. If $\sum {x}^{2}=424375$, find the confidence level.
So the idea of confidence intervals is still rather new to me and one that isn't fully clear in my mind, would someone be able to give me some hints about solving this and explain what they are doing to solve it.
I get that the confidence interval is given by
$\overline{x}-z\frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+z\frac{\sigma }{\sqrt{n}}$
and that $z={\mathrm{\Phi }}^{-1}\left(c\right)$ where c is the confidence level, however this is as far as my knowledge goes.
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altaryjny94
Step 1
To match the confidence interval with your numerical data, we get
$\overline{x}-z\frac{\sigma }{\sqrt{n}}=94.93$
and $\overline{x}+z\frac{\sigma }{\sqrt{n}}=105.07$
You are interested in the confidence level, which you could know if you get z, i.e. via the normal tail. Subtracting both equations you get
$2z\frac{\sigma }{\sqrt{n}}=105.07-94.93=10.14$
so $z=\frac{10.14}{2\frac{\sigma }{\sqrt{n}}}$
Since your sample size consists of 40 elements, then $n=40$, hence
$z=\frac{10.14}{2\frac{\sigma }{\sqrt{40}}}$
and your $\sigma$ could be estimated using an unbiased estimator
${\sigma }^{2}=\frac{\sum _{i=1}^{40}{x}_{i}^{2}-\frac{1}{n}{\left(\sum _{i=1}^{40}{x}_{i}\right)}^{2}}{n-1}=\frac{424375-\frac{1}{40}{\left(\sum _{i=1}^{40}{x}_{i}\right)}^{2}}{40-1}$
Step 2
The sum $\sum {x}_{i}$ is nothing other than $\sum _{i=1}^{40}{x}_{i}=n\overline{x}$ which could be obtained by adding the first two equations:
$2\overline{x}=105.07+94.93=200$
hence
$\overline{x}=100$
so
$\sum _{i=1}^{40}{x}_{i}=n\overline{x}=40\left(100\right)=4000$
Replacing we get
$\sigma =\sqrt{\frac{424375-\frac{1}{40}{\left(4000\right)}^{2}}{40-1}}$
So
$z=\frac{10.14}{2\frac{\sqrt{\frac{424375-\frac{1}{40}{\left(4000\right)}^{2}}{40-1}}}{\sqrt{40}}}$
###### Did you like this example?
Step 1
If $z={\mathrm{\Phi }}^{-1}\left(c\right)$, then let the $100\left(2c-1\right)\mathrm{%}$ confidence interval for $\mu$ be ${\mu }_{1}<\mu <{\mu }_{2}$.
This is because the normal distribution (which is what we are assuming here) is two-tailed, so we have that
$P\left(Xz\right)=1-c$
and we seek the region $P\left(-z due to symmetry. Here, X is just a random variable.
Step 2
Therefore, using the standard inequalities for confidence intervals, we obtain the following equalities
$\begin{array}{}\text{(1)}& \overline{x}-z\frac{\sigma }{\sqrt{n}}={\mu }_{1}\end{array}$
$\begin{array}{}\text{(2)}& \overline{x}+z\frac{\sigma }{\sqrt{n}}={\mu }_{2}.\end{array}$
Adding the two yields $\overline{x}=\left({\mu }_{1}+{\mu }_{2}\right)/2$, or that
$\begin{array}{}\text{(3)}& \sum {x}_{i}=\frac{n}{2}\left({\mu }_{1}+{\mu }_{2}\right).\end{array}$
Similarly, performing $\left(2\right)-\left(1\right)$ results in
$\begin{array}{}\text{(4)}& z\frac{\sigma }{\sqrt{n}}=\frac{{\mu }_{2}-{\mu }_{1}}{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}z=\frac{\left({\mu }_{2}-{\mu }_{1}\right)\sqrt{n}}{2\sigma }\end{array}$
but recall that the unbiased estimator for the variance is
$\begin{array}{}\text{(5)}& {\sigma }^{2}=\frac{\sum {x}_{i}^{2}-\frac{1}{n}{\left(\sum {x}_{i}\right)}^{2}}{n-1}.\end{array}$
You now have enough information to calculate z, and thus $c=\mathrm{\Phi }\left(z\right)$.